A long solenoid of diameter `0.1 m` has `2xx 10^(4)` turns per metre. At the centre of the solenoid, a coil of 100 turns and radius `0.01 m` is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to `0 A` from `4A` in `0.05` .If the resistance of the coil is `10 pi^(2) ohm`, the total charge (in `mu C` ) flowing through the coil during this time is

Answer

To solve the problem step by step, we will follow the principles of electromagnetic induction and the formulas related to solenoids and coils.

### Step 1: Identify the given values
- Diameter of the solenoid, \( d = 0.1 \, \text{m} \)
- Radius of the solenoid, \( r_s = \frac{d}{2} = 0.05 \, \text{m} \)
- Turns per unit length of the solenoid, \( n = 2 \times 10^4 \, \text{turns/m} \)
- Initial current in the solenoid, \( I_0 = 4 \, \text{A} \)
- Final current in the solenoid, \( I_f = 0 \, \text{A} \)
- Time duration, \( \Delta t = 0.05 \, \text{s} \)
- Number of turns in the coil, \( N = 100 \)
- Radius of the coil, \( r_c = 0.01 \, \text{m} \)
- Resistance of the coil, \( R = 10\pi^2 \, \Omega \)

### Step 2: Calculate the change in current
The change in current (\( \Delta I \)) is given by:
\[ \Delta I = I_f - I_0 = 0 - 4 = -4 \, \text{A} \]

### Step 3: Calculate the magnetic field inside the solenoid
The magnetic field \( B \) inside a solenoid is given by:
\[ B = \mu_0 n I \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

### Step 4: Calculate the change in magnetic field
The change in magnetic field (\( \Delta B \)) can be calculated as:
\[ \Delta B = \mu_0 n \Delta I \]
Substituting the values:
\[ \Delta B = (4\pi \times 10^{-7}) \times (2 \times 10^4) \times (-4) \]
Calculating this gives:
\[ \Delta B = -32 \times 10^{-3} \, \text{T} = -0.032 \, \text{T} \]

### Step 5: Calculate the area of the coil
The area \( A \) of the coil is given by:
\[ A = \pi r_c^2 = \pi (0.01)^2 = \pi \times 10^{-4} \, \text{m}^2 \]

### Step 6: Calculate the induced EMF (\( \mathcal{E} \))
The induced EMF can be calculated using Faraday's law:
\[ \mathcal{E} = -N \frac{\Delta B}{\Delta t} A \]
Substituting the values:
\[ \mathcal{E} = -100 \times \frac{-0.032}{0.05} \times \pi \times 10^{-4} \]
Calculating this gives:
\[ \mathcal{E} = 100 \times \frac{0.032}{0.05} \times \pi \times 10^{-4} = 64\pi \times 10^{-4} \, \text{V} \]

### Step 7: Calculate the total charge (\( Q \)) flowing through the coil
Using the formula:
\[ Q = \frac{\mathcal{E} \Delta t}{R} \]
Substituting the values:
\[ Q = \frac{64\pi \times 10^{-4} \times 0.05}{10\pi^2} \]
This simplifies to:
\[ Q = \frac{64 \times 0.05}{10\pi} \times 10^{-4} = \frac{3.2}{\pi} \times 10^{-4} \, \text{C} \]

### Step 8: Convert charge to microcoulombs
To convert to microcoulombs:
\[ Q = \frac{3.2}{\pi} \times 10^{-4} \times 10^{6} \, \mu C = \frac{320}{\pi} \, \mu C \]
Calculating this gives approximately:
\[ Q \approx 32 \, \mu C \]

### Final Answer
The total charge flowing through the coil during this time is approximately \( 32 \, \mu C \).

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